Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{5x + 15}{5x^2 + 10x - 15} \div \dfrac{3x^2 + 6x}{x^3 - 8x^2 + 7x} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{5x + 15}{5x^2 + 10x - 15} \times \dfrac{x^3 - 8x^2 + 7x}{3x^2 + 6x} $ First factor out any common factors. $z = \dfrac{5(x + 3)}{5(x^2 + 2x - 3)} \times \dfrac{x(x^2 - 8x + 7)}{3x(x + 2)} $ Then factor the quadratic expressions. $z = \dfrac {5(x + 3)} {5(x - 1)(x + 3)} \times \dfrac {x(x - 1)(x - 7)} {3x(x + 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {5(x + 3) \times x(x - 1)(x - 7) } { 5(x - 1)(x + 3) \times 3x(x + 2)} $ $z = \dfrac {5x(x - 1)(x - 7)(x + 3)} {15x(x - 1)(x + 3)(x + 2)} $ Notice that $(x - 1)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {5x\cancel{(x - 1)}(x - 7)(x + 3)} {15x\cancel{(x - 1)}(x + 3)(x + 2)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $z = \dfrac {5x\cancel{(x - 1)}(x - 7)\cancel{(x + 3)}} {15x\cancel{(x - 1)}\cancel{(x + 3)}(x + 2)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $z = \dfrac {5x(x - 7)} {15x(x + 2)} $ $ z = \dfrac{x - 7}{3(x + 2)}; x \neq 1; x \neq -3 $